Residue of pole at higher power in the denominator

Question

I need to calculate the residue of $\frac{1}{z^{2017}}$

My thought process would be to use the derivative formula for a pole of higher order for the pole at $0$ of order $2017$ but I can’t be expected to find the $2016$th derivative of this surely.

Answer 1

That residue at $0$ is $0$, of course, since$$\frac1{z^{2\,017}}=\frac0z+\frac0{z^2}+\cdots+\frac0{z^{2\,016}}+\frac1{z^{2\,017}}+\frac0{z^{2\,018}}+\cdots$$